Definition of Orthogonal Matrices

An \( n \times n \) matrix whose columns form an orthonormal set is called an orthogonal matrix.
As a reminder, a set of vectors is orthonormal if each vector is a unit vector ( length or norm of the vector is equal to \( 1\)) and each vector in the set is orthogonal to all other vectors in the set.
These are examples of orthogonal matrices.
a) \( \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & -1 & 0\\ \end{bmatrix} \)        b) \( \begin{bmatrix} \dfrac{1}{\sqrt {10}} & 0 & - \dfrac{3}{\sqrt {10}}\\ \dfrac{3}{\sqrt {10}} & 0 & \dfrac{1}{\sqrt {10}} \\ 0 & -1 & 0\\ \end{bmatrix} \)        c) \( \begin{bmatrix} \dfrac{1}{\sqrt{6}}&0&\dfrac{5}{\sqrt{30}}\\ \:\dfrac{1}{\sqrt{6}}&-\dfrac{2}{\sqrt{5}}&-\dfrac{1}{\sqrt{30}}\\ \:\dfrac{2}{\sqrt{6}}&\dfrac{1}{\sqrt{5}}&-\dfrac{2}{\sqrt{30}} \end{bmatrix} \)        d) \( \begin{bmatrix} \cos x& -\sin x \\ \sin x&\cos x \end{bmatrix} \)

Properties of Orthogonal Matrices

A list of the most important properties of orthogonal matrices is given below. If \( Q \) is an orthogonal matrix, then

1.   \( Q^{-1} = Q^T \) ; this the most important property of orthogonal matrices as the inverse is simply the transpose.
2.   the rows of \( Q \) form an orthonormal set.
3.   \( Q^{-1} \) is an orthogonal matrix
4.   det \( ( Q ) = \pm 1 \)
5.   if \( \lambda \) is an eigrnvalue of Q, then \( |\lambda| = 1 \)
6.   if \( Q_1 \) and \( Q_2 \) are \( n \times n \) orthogonal matrices, then \( Q_1 Q_2 \) is also an orthogonal matrix.

Examples with Solutions

Example 1
The matrices \( Q_1 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \) and \( Q_2 = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} \) are orthogonal. Verify that the product \( Q_1 Q_2 \) is also orthogonal (Property 6 above)

Solution
We first calculate the product \( Q_1 Q_2 \)
\( Q_1 Q_2 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 0&0&-1\\ 0&-1&0\\ 1&0&0 \end{bmatrix} \)
Let \( \textbf v_1 , \textbf v_2 , \textbf v_3 \) be the columns of the matrix \( Q_1 Q_2 \) found above.
\( \textbf {v}_1 = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \) , \( \textbf {v}_2 = \begin{bmatrix} 0\\ -1\\ 0 \end{bmatrix} \) , \( \textbf {v}_3 = \begin{bmatrix} -1\\ 0\\ 0 \end{bmatrix} \)
Let us calculate the length or norm of each column
\( || \textbf {v}_1 || = \sqrt {0^2+0^2+1^2} = 1 \)
\( || \textbf {v}_2 || = \sqrt {0^2+(-1)^2+0^2} = 1 \)
\( || \textbf {v}_3 || = \sqrt {(-1)^2+0^2+0^2} = 1 \)
All three vectors are unit vectors.
Calculate the inner product of all pairs of vectors that can be made from the vectors \( \textbf v_1 , \textbf v_2 , \textbf v_3 \)
\( \textbf {v}_1 \cdot \textbf {v}_2 = 0 \cdot 0 + 0 \cdot (-1) + 1 \cdot 0 = 0 \)
\( \textbf {v}_1 \cdot \textbf {v}_3 = 0 \cdot (-1) + 0 \cdot 0 + 1 \cdot 0 = 0 \)
\( \textbf {v}_2 \cdot \textbf {v}_3 = 0 \cdot (-1) + (-1) \cdot 0 + 0 \cdot 0 = 0 \)
The three vectors form an orthogonal set.
The three columns of the matrix \( Q_1 Q_2 \) are orthogonal and have norm or length equal to 1 and are therefore orthonormal.

Example 2
Use a calculator to find the inverse of the orthogonal matrix matrix \( Q = \begin{bmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{bmatrix} \) and verify Property 1 above.

Solution
Use any matrix calculator to find
\( Q^{-1} = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \end{bmatrix} \)
Find the transpose of matrix \( Q \)
\( Q^T = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \end{bmatrix} \)
Hence \( Q^{-1} = Q^T \) , property 1 above.

Example 3
Find the real constants \( a \) and \( b \) in the matrix \( Q = \begin{bmatrix} -1 & 0 & 0 \\ 0 & \dfrac{1}{\sqrt 2} & a \\ 0 & \dfrac{1}{\sqrt 2} & b \end{bmatrix} \) such that \( Q \) is orthogonal.

Solution
Let \( \textbf v_1 , \textbf v_2 , \textbf v_3 \) be the columns of the matrix \( Q \) given above such that
\( \textbf {v}_1 = \begin{bmatrix} -1\\ 0\\ 0 \end{bmatrix} \) , \( \textbf {v}_2 = \begin{bmatrix} 0\\ \dfrac{1}{\sqrt 2}\\ \dfrac{1}{\sqrt 2} \end{bmatrix} \) , \( \textbf {v}_3 = \begin{bmatrix} 0\\ a\\ b \end{bmatrix} \)
Two sets of conditions for matrix \( Q \) to be orthogonal:
1) The norm of each column \( \textbf v_1 , \textbf v_2 , \textbf v_3 \) must equal to 1
\( || \textbf v_1 || = \sqrt {(-1)^2+0^2+0^2} = 1 \)
\( || \textbf v_2 || = \sqrt {0^2+ \left(\dfrac{1}{\sqrt 2} \right)^2+\left(\dfrac{1}{\sqrt 2} \right)^2 } = 1 \)
\( || \textbf v_3 || = \sqrt {0^2+a^2+b^2} = \sqrt {a^2+b^2} = 1 \)         (I)
2) The inner product of any two vectors must be equal to zero (orthogonal vetcors)
\( \textbf {v}_1 \cdot \textbf {v}_2 = (-1) \cdot 0 + 0 \cdot \left(\dfrac{1}{\sqrt 2} \right) + 0 \cdot \left(\dfrac{1}{\sqrt 2} \right) = 0 \)
\( \textbf {v}_1 \cdot \textbf {v}_3 = (-1) \cdot 0 + 0 \cdot a + 0 \cdot b = 0 \)
\( \textbf {v}_2 \cdot \textbf {v}_3 = 0 \cdot 0 + \dfrac{1}{\sqrt 2} \cdot a + \dfrac{1}{\sqrt 2} \cdot b = \dfrac{1}{\sqrt 2} \cdot a + \dfrac{1}{\sqrt 2} \cdot b = 0 \)         (II)
For all conditions to be satisfied, we need to solve equations (I) and (II) above
Equation (II) above gives \( a = - b \)
Substitute \( a \) by \( - b \) in equation (I) to obtain
\( \sqrt {2 b^2 } = 1 \)
Solve to obtain
\( b = \pm \dfrac{1}{\sqrt 2} \)
Two solutions to the above question
\( a = - \dfrac{1}{\sqrt 2} \) and \( b = \dfrac{1}{\sqrt 2} \)
\( a = \dfrac{1}{\sqrt 2} \) and \( b = - \dfrac{1}{\sqrt 2} \)

Questions (with solutions given below)

• Part 1
  Matrix \( Q = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) is orthogonal. Verify the first 5 properties, listed above, for matrix \( Q \).
  
• Part 2
  1) Which of the following matrices are orthogonal?
  a) \( A = \begin{bmatrix} \dfrac{1}{\sqrt 2} & \dfrac{1}{\sqrt 2} \\ - \dfrac{1}{\sqrt 2} & \dfrac{1}{\sqrt 2} \end{bmatrix} \) , b) \( B = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \) , c) \( C = \begin{bmatrix} \dfrac{1}{\sqrt {10}} & - \dfrac{3}{\sqrt {10}} & 0 \\ 0 & 0 & 1 \\ \dfrac{3}{\sqrt {10}} & \dfrac{1}{\sqrt {10}} & 0 \end{bmatrix} \)
  
• Part 3
  1) Find all matrices of the form \( A = \begin{bmatrix} p & q\\ \dfrac{1}{\sqrt {3}} & r \end{bmatrix} \) that are orthogonal.
  
• Part 4
  Find the inverse of matrix \( A = \begin{bmatrix} 0 & \dfrac{5}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{1}{\sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{2}{\sqrt 5} & \dfrac{2}{3 \sqrt 5} & - \dfrac{1}{3} \end{bmatrix} \)
  

Solutions to the Above Questions

• Part 1
  Property 1
  Use a calculator to calculate \( Q^{-1} \)
  \( Q^{-1} = \begin{bmatrix}0&-1&0\\ 1&0&0\\ 0&0&1\end{bmatrix} \)
  Determine \( Q^T \)
  \( Q^T = \begin{bmatrix}0&-1&0\\ 1&0&0\\ 0&0&1\end{bmatrix} \)
  We conclude that \( Q^{-1} = Q^T \)
  Property 2
  The rows of \( Q \) may be represented by the vectors \( \textbf {v}_1 = \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix} \) , \( \textbf {v}_2 = \begin{bmatrix} -1\\ 0\\ 0 \end{bmatrix} \) , \( \textbf {v}_3 = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \)
  Calculate the length or norm of each vector
  \( ||\textbf {v}_1 || = \sqrt {0^2+1^2+0^2} = 1 \)
  \( ||\textbf {v}_2 || = \sqrt {(-1)^2+0^2+0^2} = 1 \)
  \( ||\textbf {v}_3 || = \sqrt {0^2+0^2+1^2} = 1 \)
  Calculate the inner product of each pair of vetors
  \( \textbf {v}_1 \cdot \textbf {v}_2 = 0\cdot(-1) + 1\cdot0 + 0\cdot0 = 0 \)
  \( \textbf {v}_1 \cdot \textbf {v}_3 = 0\cdot0 + 1\cdot0 + 0\cdot1 = 0 \)
  \( \textbf {v}_2 \cdot \textbf {v}_3 = (-1)\cdot0 + 0\cdot0 + 0\cdot1 = 0 \)
  The rows of \( Q \) form an orthonormal set of vectors.
  Property 3
  We have already calculated \( Q^{-1} = \begin{bmatrix}0&-1&0\\ 1&0&0\\ 0&0&1\end{bmatrix} \)
  We can easily check that the column vectors of \( Q^{-1} \) are the same as the row vectors \( \textbf {v}_1 , \textbf {v}_2 , \textbf {v}_3 \) already shown above (in property 2) that they form an orthonormal set. Hence \( Q^{-1} \) is an orthogonal matrix.
  Property 4
  Using the first row of \( Q \), we calculate the determinant as follows
  Det \( (Q) = - 1 \cdot \text{Det} \begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix} = 1 \)
  Property 5
  Find the eigenvalues of \( Q \) by solving the equation
  \( \det \left\{ \:\begin{bmatrix}\:\:\:0\:\:&\:1\:&\:0\:\\ \:\:\:\:\:-1\:\:\:&\:0\:&\:0\:\\ \:\:\:\:\:0\:&\:0\:&\:1\end{bmatrix} - \lambda \:\begin{bmatrix}\:\:\:1\:\:&\:0\:&\:0\:\\ \:\:\:\:\:0\:\:\:&\:1\:&\:0\:\\ \:\:\:\:\:0\:&\:0\:&\:1\end{bmatrix} \right \} =0 \)
  Simplify the left side of the above and rewrite as
  \( \det \begin{bmatrix}-\lambda&1&0\\ -1&-\lambda&0\\ 0&0&1-\lambda\end{bmatrix} = 0 \)
  Use the third row (it has 2 zeros)) to evaluate the determinant on the left side of the equation
  \( (1 - \lambda) ( \lambda^2 + 1 ) = 0 \)
  Solve the above equation to find all eigenvalues
  \( \lambda_1 = 1 \) , \( \lambda_2 = i \) and \( \lambda_3 = - i \)
  and we can now conclude that
  \( |\lambda_1| = 1 \) , \( |\lambda_2| = 1 \) and \( |\lambda_3| = 1 \)
  
  
  
• Part 2
  Matrices A and C are orthogonal but matrix B is not because its columns 2 and 3 are not orthogonal.
  
  
• Part 3
  Two conditions must be satified for matrix \( A = \begin{bmatrix} p & q\\ \dfrac{1}{\sqrt {3}} & r \end{bmatrix} \) to be orthogonal.
  1) The vectors formed by the column must be unit vectors (norm equal to 1). Hence
  \( \sqrt {p^2 + \left(\dfrac{1}{\sqrt {3}}\right)^2} = 1 \)         (I)
  and
  \( \sqrt {q^2 + r^2 } = 1 \)         (II)
  Solve equation (I) to obtain two solutions
  \( p_1=\sqrt{\dfrac{2}{3}}\) and \( p_2=-\sqrt{\dfrac{2}{3}} \)
  2) The two columns must be orthogonal which gives the equation
  \( p q + \dfrac{r}{\sqrt {3}} = 0 \)         (III)
  
  Use the first solution \( p_1 \) to find \( q\) and \( r \)
  Let \( p = \sqrt{\dfrac{2}{3}} \) and substitute in equation (III)
  \( \sqrt{\dfrac{2}{3}} q + \dfrac{r}{\sqrt {3}} = 0 \)
  which gives
  \( q = - \dfrac{\sqrt{2}r}{2} \)         (IV)
  Substitute in equation (II) to obtain the equation
  \( \sqrt { \left(- \dfrac{\sqrt{2}r}{2}\right)^2 + r^2 } = 1 \)
  Solve the above to find
  \( r_1 =\sqrt{\dfrac{2}{3}} \) and \( r_2 = -\sqrt{\dfrac{2}{3}} \)
  Substitute in equation (IV) to obtain the corresponding values of \( q \)
  \( q_1 = -\dfrac{\sqrt{3}}{3} \) and \( q_2 = \dfrac{\sqrt{3}}{3} \)
  
  Use the second solution \( p_2 \) to find \( q\) and \( r \)
  Let \( p = - \sqrt{\dfrac{2}{3}} \) and substitute in equation (III)
  \( - \sqrt{\dfrac{2}{3}} q + \dfrac{r}{\sqrt {3}} = 0 \)
  which gives
  \( q = \dfrac{\sqrt{2}r}{2} \)         (V)
  Substitute in equation (II) to obtain the equation
  \( \sqrt {\left(\dfrac{\sqrt{2}r}{2}\right)^2 + r^2 } = 1 \)
  Solve the above to find
  \( r_3 =\sqrt{\dfrac{2}{3}} \) and \( r_4 = -\sqrt{\dfrac{2}{3}} \)
  Substitute in equation (IV) to obtain the corresponding values of \( q \)
  \( q_3 = \dfrac{\sqrt{3}}{3} \) and \( q_4 = -\dfrac{\sqrt{3}}{3} \)
  
  The matrices are
  \(\begin{bmatrix} \sqrt{\dfrac{2}{3}} & -\dfrac{\sqrt{3}}{3}\\ \dfrac{1}{\sqrt {3}} & \sqrt{\dfrac{2}{3}} \end{bmatrix} \) , \(\begin{bmatrix} \sqrt{\dfrac{2}{3}} & \dfrac{\sqrt{3}}{3}\\ \dfrac{1}{\sqrt {3}} & -\sqrt{\dfrac{2}{3}} \end{bmatrix} \) , \(\begin{bmatrix} - \sqrt{\dfrac{2}{3}} & \dfrac{\sqrt{3}}{3}\\ \dfrac{1}{\sqrt {3}} & \sqrt{\dfrac{2}{3}} \end{bmatrix} \) , \(\begin{bmatrix} - \sqrt{\dfrac{2}{3}} & -\dfrac{\sqrt{3}}{3}\\ \dfrac{1}{\sqrt {3}} & -\sqrt{\dfrac{2}{3}} \end{bmatrix} \) ,
  
  
• Part 4
  Given matrix \( A = \begin{bmatrix} 0 & \dfrac{5}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{1}{\sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{2}{\sqrt 5} & \dfrac{2}{3 \sqrt 5} & - \dfrac{1}{3} \end{bmatrix} \)
  Any of the known method may be used to find the inverse of matrix \( A \). But a close examination of the matrix reveals that the matrix is orthogonal.
  Let us prove that the given matrix is orthogonal.
  Let \( \textbf v_1 , \textbf v_2 , \textbf v_3 \) be the columns of matrix \( A \) given above such that
  \( \textbf {v}_1 = \begin{bmatrix} 0\\ \dfrac{1}{\sqrt 5} \\ \dfrac{2}{\sqrt 5} \end{bmatrix} \) , \( \textbf {v}_2 = \begin{bmatrix} \dfrac{5}{3 \sqrt 5}\\ - \dfrac{4}{3 \sqrt 5}\\ \dfrac{2}{3 \sqrt 5} \end{bmatrix} \) , \( \textbf {v}_3 = \begin{bmatrix} \dfrac{2}{3}\\ \dfrac{2}{3}\\ -\dfrac{1}{3} \end{bmatrix} \)
  Let us calculate the length or norm of each vector.
  \( || \textbf {v}_1 || = \sqrt {0^2+ \left(\dfrac{1}{\sqrt 5} \right)^2+ \left(\dfrac{2}{\sqrt 5} \right)^2} = 1 \)
  \( || \textbf {v}_2 || = \sqrt { \left(\dfrac{5}{3 \sqrt 5}\right)^2+ \left(- \dfrac{4}{3 \sqrt 5} \right)^2+ \left(\dfrac{2}{3 \sqrt 5} \right)^2} = 1 \)
  \( || \textbf {v}_3 || = \sqrt { \left(\dfrac{2}{3} \right)^2+ \left(\dfrac{2}{3} \right)^2+ \left(-\dfrac{1}{3} \right)^2 } = 1 \)
  All three vectors formed by the columns of the given matrix are unit vectors (norm or length = 1)
  Calculate the inner product of all pairs of vectors that can be made from the vectors \( \textbf v_1 , \textbf v_2 , \textbf v_3 \)
  \( \textbf {v}_1 \cdot \textbf {v}_2 = 0 \cdot \left(\dfrac{5}{3 \sqrt 5}\right) + \left(\dfrac{1}{\sqrt 5} \right) \cdot \left(- \dfrac{4}{3 \sqrt 5} \right) + \left(\dfrac{2}{\sqrt 5} \right) \cdot \left(\dfrac{2}{3 \sqrt 5} \right) = 0 \)
  \( \textbf {v}_1 \cdot \textbf {v}_3 = 0 \cdot \left(\dfrac{2}{3}\right) + \left(\dfrac{1}{\sqrt 5} \right) \cdot \left(\dfrac{2}{3}\right) + \left(\dfrac{2}{\sqrt 5} \right) \cdot \left(-\dfrac{1}{3}\right) = 0 \)
  \( \textbf {v}_2 \cdot \textbf {v}_3 = \left(\dfrac{5}{3 \sqrt 5}\right) \cdot \left(\dfrac{2}{3}\right) + \left(- \dfrac{4}{3 \sqrt 5} \right) \cdot \left(\dfrac{2}{3}\right) + \left(\dfrac{2}{3 \sqrt 5} \right) \cdot \left(-\dfrac{1}{3}\right) = 0 \)
  The three vectors formed by the columns of the given matrix are orthogonal. Therefore the given matrix is orthogonal and we may the propetry \( A^{-1} = A^T \); hence the inverse of matrix \( A \) is given by the transpose of matrix \( A \).
  \( A^{-1} = \begin{bmatrix} 0 & \dfrac{1}{\sqrt 5} & \dfrac{2}{\sqrt 5} \\ \dfrac{5}{3 \sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3 \sqrt 5} \\ \dfrac{2}{3} & \dfrac{2}{3} & - \dfrac{1}{3} \end{bmatrix} \)
  

More References and links

• Matrices with Examples and Questions with Solutions.
  
• Determinant of a Square Matrix.
  
• Inverse Matrix Questions with Solutions.
  
• Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres
• Introduction to Linear Algebra - Fifth Edition (2016) - Gilbert Strang
• Linear Algebra Done Right - third edition, 2015 - Sheldon Axler
• Linear Algebra with Applications - 2012 - Gareth Williams